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Nimfa-mama [501]

3 years ago

14

Describe the Earth's layers in relation to density

1 answer:

jok3333 [9.3K]3 years ago

7 0

<span>Our Earth is structured around the densities of the materials which make it up.One property of density is that it determines the way materials in a mixture are sorted. This property of matter results in the layering and structure of Earth's atmosphere, water, crust, and interior.</span>

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Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole

trapecia [35]

Answer : The enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as $\Delta H^o$

The equation used to calculate enthalpy change is of a reaction is:

$\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]$

The equilibrium reaction follows:

$C_6H_{12}O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(n_{(CO_2)}\times \Delta H^o_f_{(CO_2)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(C_6H_{12}O_6)}\times \Delta H^o_f_{(C_6H_{12}O_6)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]$

We are given:

$\Delta H^o_f_{(C_6H_{12}O_6(s))}=-1260kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(CO_2(g))}=-393.5kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.8kJ/mol$

Putting values in above equation, we get:

$\Delta H^o_{rxn}=[(6\times -393.5)+(6\times -285.8)]-[(1\times -1260)+(6\times 0)]=-2815.8kJ/mol$

Therefore, the enthalpy of combustion per mole of $C_6H_{12}O_6$ is -2815.8 kJ/mol

6 0

4 years ago

Read 2 more answers

20. Which of the following is a method of purifying water? A. Adding iodine B. Freezing C. Aeration D. Fixation

Oxana [17]

The answer is A, adding Iodine

7 0

3 years ago

Read 2 more answers

Islands have cool and calm weather throughout the year? Why?

saveliy_v [14]

There’s no one around and the breeze from the Ocean makes it cool

8 0

3 years ago

A 5.098 mass % aqueous solution of potassium hydroxide has a density of 1.05 g/mL. Calculate the molality of the solution. Give

GarryVolchara [31]

<u>Answer:</u> The molarity of solution is 0.954 M

<u>Explanation:</u>

We are given:

5.098 mass % solution of potassium hydroxide

This means that 5.098 grams of potassium hydroxide is present in 100 grams of solution

To calculate volume of a substance, we use the equation:

$\text{Density of substance}=\frac{\text{Mass of substance}}{\text{Volume of substance}}$

Density of solution = 1.05 g/mL

Mass of solution = 100 g

Putting values in above equation, we get:

$1.05g/mL=\frac{100g}{\text{Volume of solution}}\\\\\text{Volume of solution}=\frac{100g}{1.05g/mL}=95.24mL$

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of potassium hydroxide = 5.098 g

Molar mass of potassium hydroxide = 56.1 g/mol

Volume of solution = 95.24 mL

Putting values in above equation, we get:

$\text{Molarity of solution}=\frac{5.098\times 1000}{56.1\times 95.24}\\\\\text{Molarity of solution}=0.954M$

Hence, the molarity of solution is 0.954 M

7 0

4 years ago

Pls help me ASAP thank you!!

Stolb23 [73]

Sorry i cant even read that

7 0

3 years ago